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Question Number 112454 by bemath last updated on 08/Sep/20
(1) find the locus ∣z−z_1 ∣ = 2 meets the positive real axis (2)On a single Argand diagram, sketch the loci → { ((∣z−z_1 ∣=2)),((arg(z−z_2 )=(π/4))) :}
$$\left(\mathrm{1}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\:\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid\:=\:\mathrm{2}\:\mathrm{meets} \\ $$$$\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\left(\mathrm{2}\right)\mathrm{On}\:\mathrm{a}\:\mathrm{single}\:\mathrm{Argand}\:\mathrm{diagram},\:\mathrm{sketch} \\ $$$$\mathrm{the}\:\mathrm{loci}\:\rightarrow\begin{cases}{\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid=\mathrm{2}}\\{\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$
Commented by MJS_new last updated on 08/Sep/20
(2) we cannot scetch the loci if we don′t know any coordinates. ∣z−z_1 ∣=2 is the equations of all circles with radius 2 arg (z−z_2 ) =(π/4) is any straight line with slope 45° let z−z_2 =re^(iθ) ⇒ arg re^(iθ) =θ ⇒ θ=(π/4) arg ((x−p)+(y−q)i) =(π/4) leads to y=x−p+q
$$\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{scetch}\:\mathrm{the}\:\mathrm{loci}\:\mathrm{if}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know} \\ $$$$\mathrm{any}\:\mathrm{coordinates}. \\ $$$$\mid{z}−{z}_{\mathrm{1}} \mid=\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{all}\:\mathrm{circles}\:\mathrm{with} \\ $$$$\mathrm{radius}\:\mathrm{2} \\ $$$$\mathrm{arg}\:\left({z}−{z}_{\mathrm{2}} \right)\:=\frac{\pi}{\mathrm{4}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{slope}\:\mathrm{45}° \\ $$$$\mathrm{let}\:{z}−{z}_{\mathrm{2}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\mathrm{arg}\:{r}\mathrm{e}^{\mathrm{i}\theta} \:=\theta\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{arg}\:\left(\left({x}−{p}\right)+\left({y}−{q}\right)\mathrm{i}\right)\:=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{leads}\:\mathrm{to}\:{y}={x}−{p}+{q} \\ $$
Answered by MJS_new last updated on 08/Sep/20
(1) assuming z_1 is given and z is variable we have z_1 =p+qi ∧ z=x+yi ∣z−z_1 ∣=2 ∣(x−p)+(y−q)i∣=2 (√((x−p)^2 +(y−q)^2 ))=2 both sides are ≥0 ⇒ we are allowed to square (x−p)^2 +(y−q)^2 =4 which is a circle with center ((p),(q) ) and radius 4 if it meets the positive real axis depends on the values if p and q −2≤q≤2 ⇒ it meets the real axis p>2 ⇒ it meets the positive real axis we find the intersection point(s) putting y=0 (x−p)^2 +q^2 =4 ⇒ x=p±(√(4−q^2 )) with p>2∧−2≤q≤2
$$\left(\mathrm{1}\right) \\ $$$$\mathrm{assuming}\:{z}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{given}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{variable}\:\mathrm{we}\:\mathrm{have} \\ $$$${z}_{\mathrm{1}} ={p}+{q}\mathrm{i}\:\wedge\:{z}={x}+{y}\mathrm{i} \\ $$$$\mid{z}−{z}_{\mathrm{1}} \mid=\mathrm{2} \\ $$$$\mid\left({x}−{p}\right)+\left({y}−{q}\right)\mathrm{i}\mid=\mathrm{2} \\ $$$$\sqrt{\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:\mathrm{and}\:\mathrm{radius}\:\mathrm{4} \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis}\:\mathrm{depends}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{values}\:\mathrm{if}\:{p}\:\mathrm{and}\:{q} \\ $$$$−\mathrm{2}\leqslant{q}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{real}\:\mathrm{axis} \\ $$$${p}>\mathrm{2}\:\Rightarrow\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point}\left(\mathrm{s}\right)\:\mathrm{putting} \\ $$$${y}=\mathrm{0} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\:{x}={p}\pm\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\:\mathrm{with}\:{p}>\mathrm{2}\wedge−\mathrm{2}\leqslant{q}\leqslant\mathrm{2} \\ $$
Commented by bemath last updated on 08/Sep/20
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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